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Ideal Gas Equation

60 mL of a mi...

Question

$60$ mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If $38$ mL of $N_{2}$ was formed, then the volume (in mL) of nitric oxide in the mixture is (all measurements are made at constant $P$ and $T$ ) :

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Solution

Let the volume of $N O a n d N_{2} O$ be $a$ and $b$ mL respectively.
Then, $a + b = 60 . . . (i)$
$N O + H_{2} \to \frac{1}{2} N_{2} + H_{2} O$
$a$ mL $0$
$0$ $\frac{a}{2}$
$N_{2} O + H_{2} \to N_{2} + H_{2} O$
$b$ mL $0$
$0$ $b$
$∴$ Total $N_{2}$ formed $= (\frac{a}{2}) + b = 38$ ...(ii)
By equations (i) and (ii), $a = 44$ mL, $b = 16$ mL.
So, volume of nitric oxide $N O = 44$ mL.

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